## Precalculus (6th Edition) Blitzer

$\{1, 7\}$
$\displaystyle \frac{3}{x-3}+\frac{5}{x-4}=\frac{x^{2}-20}{x^{2}-7x+12}$ The denominator on the RHS can be factored: Two factors of 12 that add to -7 are -4 and -3 $\displaystyle \frac{3}{x-3}+\frac{5}{x-4}=\frac{x^{2}-20}{(x-3)(x-4)} \qquad$ ... multiply with LCD = $(x-3)(x-4)$ We want to write the equation as $ax^{2}+bx+c=0$ First, we exclude any x that yields 0 in a denominator $x\neq 3,4$ $3(x-4)+5(x-3)=x^{2}-20$ $3x-12+5x-15=x^{2}-20$ $8x-17==x^{2}-20$ $0=x^{2}-8x+7$ We factor this trinomial by finding two factors of $7$ that add to $-8$; they are $-7$ and $-1$. $(x-7)(x-1)=0$ $x=7,\qquad x=1$ The solution set is $\{1, 7\}.$